Sed remove everything between first and second instance of a character?

Question

I'm not very experienced with sed, but If I have a string like this:

asdf | this is something | something else | nothing | qwerty

can I remove everything between the first and second instances of |, plus one of them?

The ideal output would be:

asdf | something else | nothing | qwerty

I tried sed 's/|*|//2', but this only removes the second pipe.

Thanks

iruvar · Accepted Answer · 2013-08-29 21:10:10Z

s/|[^|]*|/|/ should do the job

echo 'asdf | this is something | something else | nothing | qwerty' | sed 's/|[^|]*|/|/' asdf | something else | nothing | qwerty

anubhava · Accepted Answer · 2013-08-29 21:25:57Z

Can be done using awk also:

awk -F '|' -v OFS='|' '{sub($2 " *\\|", "")}1' <<< "$str" asdf | something else | nothing | qwerty

Using pure BASH:

echo "${str%%|*}|${str#*|*|}" asdf | something else | nothing | qwerty

Kent · Accepted Answer · 2013-08-29 21:12:06Z

check this:

sed 's/|[^|]*//'

with your example

kent$ sed 's/|[^|]*//' <<<"asdf | this is something | something else | nothing | qwerty" asdf | something else | nothing | qwerty

3 Answers 3